Question

Scoring Scheme: 3-3-2-1 Part III. For each trial, enter the amount of heat gained by the chemical system of ammonium nitrate, qrxn. The specific heat of water is 4.184 J/goC. Report your answer using 4 digits. Trial

Answer 1

Question:

The information given is:

Trial #      Tiwater     T f           ΔT        Masswater  (m)

       #1:     21.2         10.8        10.8       24.990

       #2:    20.8        9.50       9.5         25.000

       #3:    20.9        9.20       9.2         25.010

Answer:

The heat of the reaction is -5985 J

Explanation:

The heat absorbed by the water is given by

ΔQ = m·c·ΔT  

From which

∑ (ΔT·m)/3 = 278.34 kg·°C

ΔQ = c×∑ (ΔT·m)/3 = 4.184 J/g·°C×278.34 kg·°C = 1164.565 J

ΔQ Calorimter = Specific heat capacity of calorimeter, $c_{calorimeter}$ × ΔT$_{average}$

Where the $c_{calorimeter}$ = 443 J/°C for example, we have

ΔQ Calorimter = 443×11.133 = 4820.733 J

From which the heat of reaction is then

$\Delta Q_{reaction} = -(\Delta Q_{water} + \Delta Q_{calorimter})$

$\Delta Q_{reaction} = -5985.298 \, J$

Using 4 digits, we get

$\Delta Q_{reaction} \approx -5985 \, J$.